g(x)= 1/10(x+1)^2( x-3)^3
Vertical asymptote occur when denominator is zero.
So no vertical assymptote. Degree is x^5 , odd and
leading coefficient is (+) , so end behavior is "down /up"
i.e y -> -oo as x -> -oo and
y -> oo as x -> oo
Zeros are x= -1 with multiplicity of 2 and
x= 3 with multiplicity of 3:. x intercepts are
x=-1 and x=3, y intercept: Putting x=0 in the equation
we get,y= 1/10*(1)^2* (-3)^3 = -2.7
g(x)= 1/10(x+1)^2( x-3)^3
g^'(x)= 1/10[2(x+1)(x-3)^3+(x+1)^2*3(x-3)^2] or
g^'(x)= 1/10[(x+1)(x-3)^2(2(x-3)+3(x+1)] or
g^'(x)= 1/10[(x+1)(x-3)^2(5 x-3)]
g'(x)=0 :. slope is zero at x=-1,x=3 and x=3/5=0.6
g(0.6)~~-3.5389~~ -3.54 , critical point x=0.6
Slope at x<-0.6 , g^'(x) <0 i.e decreasing and
slope at x> -0.6 , g^'(x) >0 i.e increasing.
So (0.6, -3.54) is local minimum.
At x=-1 the graph will not cross x axis as the multiplicity
is even 2 and at x=3 the graph will cross xaxis as the
multiplicity is odd 2.
graph{1/10(x+1)^2(x-3)^3 [-10, 10, -5, 5]}[Ans]
